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I consider an electron (charge $-e$) in $x=0$ and a constant electric field $E(x) \equiv E $. If the electron has initial velocity $v_0$ with the same direction of $E$, then its potential energy is $$ U(x) = -eV(x) = -e E x $$ The total energy for $x=0$ is $$ K(0) + U(0) = \dfrac{1}{2} m v_0^2 $$ Now if I try to obtain the point where the electron will eventually stop and begin moving in the opposite direction I obtain from the energy conservation $$ \dfrac{1}{2} m v_0^2 = K(0) + U(0) = K(z) + U(z) = -e E z $$ $$ z = -\dfrac{m v_0^2}{2eE} $$ But if $E > 0$ I obtain $z<0$, while $z$ should be $z>0$ since $v_0$ has the same direction as $E$!

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Is it true that V(x) = Ex? – Alfred Centauri Apr 14 '13 at 11:45
up vote 0 down vote accepted

Firstly, you can't just assert that $V=Ex$. Absolute potential isn't defined here, since there is an infinite increase of potential energy when going from $x=-\infty$ to $x=+\infty$. We can only use absolute potential when the assertion that absolute potential is $0$ at infinity holds. However, this wasn't really your issue here, it would have worked regardless. You basically forgot a negative sign while calculating V: $$\Delta V=\color{red}{-}\int\vec E\cdot d\vec l$$

So, we use:

$$\Delta KE+\Delta PE=0$$

And, $$\Delta PE=(-e)\Delta V=(-e)(-\int\vec E\cdot d\vec l)=eE\Delta x$$

$$\frac12m0^2-\frac12mv_0^2+eE\Delta x=0$$ $$\implies eEx=\frac12mv_0^2\implies x=\frac{mv_0^2}{2E}$$

Which solves the problem.

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